Complex conjugate question?

Question

Why when you multiply a complex nbr by its conjugate do you need absolute value signs? i.e

u*u(c) = |u|^2 - where u(c) is the complex conjugate of u (because I can't figur how to get the bar on top of the u!).

Obviously u*u=u^2, but what's the point of the abs value since a squared number is not negative?

I might be missing something blindingly obvious here......

Answer 1

When u is complex, u*u = u^2 is ALSO complex, in general. (It could also be real but be negative.) So it's NOT the same as |u|^2. So the absolute value or "mod"sign is absolutely necessary in order to make the distinction between the two different things clear.

(The fact that you can write some product as u^2 DOESN'T mean "it's a squared number and so can't be negative." For example, to give a trivial example, if u were ' i,' the imaginary unit for which i^2 = - 1, then obviously that is NOT a positive number!)

This is also connected with various ways of representing complex numbers. For example, if you know about the Argand diagram representation and express u as

u = r e^(iθ),

its complex conjugate is u(c) = r e^(- iθ), so that

u*u(c) = r^2. But u*u = r^2 e^(2iθ),

which is in general quite different.

Notice also that u*u(c), being r^2, is precisely what is MEANT by the square of the MODULUS of a complex number in the Argand diagram.

If you express u instead as

u = x + i y,

then by Pythagoras's Theorem in that same Argand diagram, the value of u*u(c), that is r^2, is simply x^2+ y^2.

You can check that this is consistent by noting that in this representation, u*u(c) = (x + iy)(x - iy), which by the "difference of two squares formula" (with the second term pure imaginary!) is x^2 - (iy)^2 = x^2 + y^2.

Meanwhile, in this latter representation,

u*u = (x + iy)^2 = x^2 - y^2 + 2i xy

--- something completely different!

Live long and prosper.

Answer 2

complex aspect. try searching into google and yahoo. it could help!

Answer 3

Z = x + iy
conj Z = x - iy
Z ( conj Z) = (x + i y) (x - i y)
= x ² - i x y + i x y - i ² y ²
= x ² + y ²